Reinforcement Detailing of Beams from ETABS output

QUESTION:-

Detail design of beams in Etabs 2018.
The ETABS file for a G+4 building is provided. Run the analysis and design the RCC Moment ResistingFrame. The following challenge deals specifically with two continuous beams in the 1st floor.

1) 3 span continuous beam along grid A
2) 7 span continuous beam along grid 3

Address the following issues thereafter.
• Provide details of longitudinal and shear reinforcement for the two continuous beams. Sketch the beam elevation details or draft it using autocad software. A sample of how beam elevation details is prepared in consultancy firms is provided below. Participants can prepare it as per their preference:

• Calculate value of the maximum shear force in any one of the spans in the continuous beams, as per clause 6.3.3 (b) in IS 13920 – 2016. Also, confirm these values from shear force demand calculated by ETABS.Please note that if longitudinal reinforcement provided is more than required (as per ETABS results), the shear force demand will vary from what is provided by ETABS.

ANSWER:-

Aim:- To design the beams in detail from the Etabs 2018.

To provide details of longitudinal and shear reinforcement for the given two continuous beams.

To calculate the value of maximum shear force in any one of the spans in the continuous beams.

Introduction:-

• Beams are horizontal structural elements used for the supporting lateral loads.
• In conventional reinforced concrete structures, beams usually receive load from the floor slab, but may also be subjected to other loads such as wall loads, finishes, services installation, etc.
• The design and detailing of RC beams involves the selection of the proper beam size and the quantity of longitudinal and shear reinforcement that will satisfy ultimate and serviceability limit state requirements.
• Afterwards, It is very important that the beam reinforcements are placed and arranged properly on site, to avoid construction error.

Procedure:-

Step:-1

• First we have to open the challange 7 page and download the challenge.
• After downloading the Challenge open it in the ETABS 2020, as shown below.

STEP 2

• GO TO FIRST FLOOR PLAN
• So that click on plan tab
• one dialogbox is open
• select the first floor

• click on APPLY and then OK .
• 

Step:-3

• Next we have to check the load combinations, as shown below.

STEP :-4

• RUN ANALYSYS :

• After analysis got completed then we get maximum and minimum deflection at bottom size left cornor shown in above picture .

STEP :-5

• Click on concrete frame design tab

Step:-6

• Next we have to start the beam design by selecting the 'Start Design/Check' option, as shown below.

Step:-7

• After the completion of the beam design, we need to select the 'Display Design Info' option, as shown below.

• 

• After selecting the 'Display Design Info' option the new tab appears, In that we select the Longitudinal Reinforcing in the Design Output and click on the OK button, as shown below.

• 

• Next the longitudinal reinforcement details will appear, as shown below.

• 

STEP :-8

• CHECK THE SHERE RENFORCING :
• So that click on Concrete frame design dropdown

• click on Display Design Info tab
• then again new dialogbox is open
• shown below
• select the share reinforcing option

• click on APPLY and then click ok
• 

STEP :-9

• CHECK THE DETAILS OF PERTICULAR ELEMENT
• So that, rignt click on any element
• then one dialogbox is open

• click on Details tab
• click on Shear details

STEP-:10

Select on grid A for three span

• Next we need to draw the Reinforcement diagram of 3 span continous beam along Grid-A, and also need to provide the reinforcement details.

Reinforcement details along Grid A

• Max Ast at bottom support =1927mm²
• Max Ast at Top support = 2027 mm²

• Reinforcement Calculation
• For Top & Bottom Max Ast is 2027 and 1927 mm²
• So we can provide 2 bars of diameter 32 mm and
• Additional bar of 25 mm diameter to satisfies the overall Ast
• 2 x T32 = 1608 mm²
• 1 x T25 = 490 mm² ................... Hence Satisfy
• 2 x T32 = 1608 mm²
• 1 x T25 = 490 mm² ................... Hence Satisfy.

• For Shear details

• Right Click on Beam and Click on Details
• Click on Shear Details

• As per Clause 6.3.3 we have to provide Spacing till 2d from the face of the column is 100.
• It should not be more than that
• So provide 3 legged stirrups of 10mm diameter with 100mm spacing throught the beam length.

STEP :-11

Select on grid 3 as 7 span

• Next we need to draw the Reinforcement diagram of 7 span continous beam along Grid-3, and also need to provide the reinforcement details.

Reinforcement details along Grid 3

Max A, at bottom support = 1501 mm²

Max At at Top support = 1940 mm²

Reinforcement Calculation

For Top & Bottom Max Ast is 1940 and 1501 mm²

So we can provide 2 bars of diameter 32 mm and

Additional 1 bars of 25 mm at top to satisfies the overall Ast

2 x T32 = 1608 mm²

1 x T25 = 490 mm² ...............Hence Satisfy.

For Shear details

Right Click on and Click on Details

Click on Shear Details

• As per Clause 6.3.3 we have to provide Spacing till 2d from the face of the column is 100.
• It should not be more than that
• so provide 3 legged stirrups of 8mm diameter with 100mm spacing throughout the beam length.

STEP :-12

Now we have to do Check for Sway shear.

A] For Right Sway

We know that the formula to calcuate

From Etabs Model results substitute the value -

V_u^D+L =92 Kn

M_as = 176 kn.m

M_bh = 295 kn

Lab = 4.725 m

Put this value in above eqn we get,

V_u,a = 48 Kn

V_u,b = 230 Kn

B] For Left Sway

Follow the same process

We know the formula

From Etabs Model results substitute the value

V_u^D+L = 74 Kn

M_ah = 236 kn

M_bh = 150 kn

Lab = 4.725 m

Put this value in above eqn we get,

V_u,a = 190 Kn

V_u,b = 50 Kn

C] Existing Shear capacity in beam as per Is 456-2002

We know the formula

Put the values

Fy = 415 N/mm2

Asv =151 mm2

Spacing = 100 mm

d = 450 mm

put this value in above equection , we get

V_us) = 246 … > 230 kn (Shear sway) .............(Hence Safe)

Provide reasons for failure of the middle span along grid A. What are the possible ways this issue can be resolved?

a. The main reason of failure of middle span along grid A is shear stress is exceeded the limiting value

b. This may be because of the higher Shear force acting on that beam and due to provision of weak shear reinforcement at the end support

This Issue can be solved by

• By providing Higher grade of concrete in that particular section
• Or by reducing the Shear force in that beam section
• or by providing higher shear reinforcement which can resist the shear force in that section

Calculate value of the maximum shear force in any one of the spans in both the continuous beams, as per clause 6.3.3 (b) in IS 13920 2016.

As we have already calculated the value of shear force as per IS 13920 - 2016 Clause 6.3.3 (b)

We get the Shear sway value Right and Left support within limit

246 kn > 230 kn................... Safe).

Results

• Beam Design is done successfully with detailed reinforcement
• Provided the details of longitudinal and shear reinforcement for the two continuous beams.
• And also calculated the value of maximum shear force in the continuous beams using the Etabs software.